Chapter 15 Gen Chem 2

Introduction to Chapter 15: Equilibria of Other Reaction Classes

15.1 Precipitation and Dissolution

Mineral Nuorite

• Chemical Formula: CaF2

• Properties: Used as a semiprecious stone in jewelry due to its appearance.

• Formation Process: Hydrothermal precipitation where calcium and fluoride ions in groundwater precipitate as nuorite due to changed conditions (e.g., lower temperature).

• Influence of pH: Because fluoride ion is a weak base, its solubility is affected by pH changes in groundwater.

Learning Objectives

• Write chemical equations and equilibrium expressions for solubility equilibria.

• Perform equilibrium computations related to solubility and solute concentrations.

Solubility Equilibria

• Established when the rates of dissolution and precipitation are equal.

• Applications: Understanding solubility is essential for processes like tooth decay and water purification.

The Solubility Product (Ksp)

• Definition: The equilibrium constant for solubility equilibria, expressed as Ksp.

• Heterogeneous Mixture: Involves a saturated solution with undissolved solute.

• Example: Silver chloride (AgCl) establishes an equilibrium:

  • $ext{AgCl (s) } \rightleftharpoons ext{Ag}^+ (aq) + ext{Cl}^- (aq)$

  • The Ksp expression excludes the solid: $K_{sp} = [ ext{Ag}^+][ ext{Cl}^-]$

• Ksp Values: Values for sparingly soluble compounds can be found in corresponding appendices.

Example Problems: Writing Equations and Solubility Products

• Example 15.1: Write equations for slightly soluble ionic compounds:

  • (a) AgI dissolution equation: $ext{AgI (s) } \rightleftharpoons ext{Ag}^+ (aq) + ext{I}^- (aq)$

  • Ksp for AgI: $K_{sp} = [ ext{Ag}^+][ ext{I}^-]$

• Exercise: Write dissolution equation and Ksp for BaSO4, Ag2SO4, Al(OH)3, Pb(OH)Cl.

Ksp and Solubility Relationships

• Ksp can be related to solubility if dissolution involves only dissociation and solvation.

• Example 15.2: Calculating Ksp from CaF2 solubility:

  • Dissolution: $ext{CaF}_2 (s) \rightleftharpoons ext{Ca}^{2+} (aq) + 2 ext{F}^- (aq)$

  • Given [Ca²⁺] = $2.15 imes 10^{-4} ext{ M}$, find Ksp:

  • $K_{sp} = [Ca^{2+}][F^-]^2 = (2.15 imes 10^{-4})(2 imes 2.15 imes 10^{-4})^2 = 8.99 imes 10^{-12}$

Example Problems: Determination of Molar Solubility from Ksp

• Example 15.3: Given Ksp for CuBr (6.3 * 10–9), find molar solubility:

  • $[Cu^+] = [Br^-] = x$, yield $K_{sp} = (x)(x) = x^2$.

• Solve for x: $x = 7.9 imes 10^{-5} ext{ M}$.

Calculating Solubility Products from Gram Solubility

• Example 15.5: Finding Ksp from solubility in g/L for PbCrO4 (4.6 * 10⁻⁶ g/L):

  • First, convert to molarity: 4.6 * 10⁻⁶ g/L => 1.4 * 10⁻⁷ M.

  • $K<em>{sp} = [Pb^{2+}][CrO</em>4^{2-}] = (1.4 imes 10^{-7})(1.4 imes 10^{-7}) = 2.0 imes 10^{-14}$.

Predicting Precipitation

• Comparing Reaction Quotient (Qsp) to Ksp:

  • If $Q<em>{sp} < K</em>{sp}$: Forward reaction proceeds (no precipitation).

  • If $Q<em>{sp} > K</em>{sp}$: Reverse reaction proceeds (precipitation occurs).

Example Problem on Precipitation

• Example 15.8: Precipitation of AgCl from given concentrations in equal volumes:

  • If mixed: concentration of Ag⁺ and Cl⁻ is halved. $( K_{sp} = 1.6 imes 10^{-10})$, comparison leads to precipitation of AgCl.

Common Ion Effect

• Definition: The presence of a common ion decreases solubility.

• Example: Adding NaCl decreases AgCl solubility due to the presence of Cl⁻.

• Example Problem: Calculate the change in solubility when NaCl is added to AgCl solution.

Summary Equations

Ksp Definition

• General solubility product expression:

  • $K_{sp} = [M^m][X^n]$ for a solid $MxYy$ with solubility of ions.

Conclusion

• Understanding precipitation and dissolution is crucial for applications in environmental science, materials chemistry, and medicine. Through various examples and problems, students can learn how to calculate and predict solubilities based on equilibrium concepts.